More efficient way to factor quadratic polynomials with a coefficient?

allyourbass2212 · #1 $Old$ July 4th, 2009, 09:44 AM

I find it more so of a time sink than actually learning anything when factoring quadratic polynomials with a coefficient attached to $x^2$

For instance, for any given expression $Ax^2+bx+c$ we must find the factors of both A and C. Such is the case in the following expression
$4x^2-4x-15$
combinations to try
a.(4x+15)(x-1) b.(4x-15)(x+1)
c.(4x-1)(x+15) d.(4x+1)(x-15)
e.(4x+5)(x-3) f.(4x-5)(x+3)
g.(4x+3)(x-5) h.(4x-3)(x+5)
i.(2x+15)(2x-1) j.(2x-15)(2x+1)
k.(2x+5)(2x-3) l.(2x-5)(2x+3)

L. Is correct

This seems very long and extraneous, is there a more effective manner to factor such problems instead of having to produce a list of all possible combinations? I have a ti83 program that can do this for me, but I would rather not resort to it if it will hurt my math skills later down the road.

Many Thanks

Moo · #2 $Old$ July 4th, 2009, 09:51 AM

Hello,

Factor A : $A\left(x^2+\frac bA x+\frac cA\right)$

Then apply methods you know (completing the square, ...)

allyourbass2212 · #3 $Old$ July 4th, 2009, 10:18 AM

Quote:

Originally Posted by Moo $View Post$

Hello,

Factor A : $A\left(x^2+\frac bA x+\frac cA\right)$

Then apply methods you know (completing the square, ...)

Unfortunately Moo I am teaching myself, and not only do that I fail to see that method in the current chapter, but in future chapters as well. Therefore I am afraid I am at somewhat of a loss in understanding this method.

skeeter · #4 $Old$ July 4th, 2009, 11:03 AM

link to a video showing another factoring technique ...

http://www.revver.com/video/644652/f...g-coefficient/

Stroodle · #5 $Old$ July 4th, 2009, 11:19 AM

You can use the crisscross type method. Putting factors of 4 in the first column and factors of -15 in the second, then multiplying the top left by the bottom right, and bottom left by the top right. Then if the sum of these results is the same as the coefficient of the x term then you have your factors.
e.g.

$2\ \ -5$

$2\ \ \ \ \ 3$

$=6+ -10=-4$

It's still trial and error, but you'll find that the more of these you do, the faster and easier they become

allyourbass2212 · #6 $Old$ July 4th, 2009, 12:20 PM

Quote:

Originally Posted by skeeter $View Post$

link to a video showing another factoring technique ...

Factor a Quadratic Equation with a leading coefficient by yourtutoronlin -- Revver Online Video Sharing Network

VERY helpful thank you so very much, and everyone else too!

masters · #7 $Old$ July 4th, 2009, 01:15 PM

Hi Allyourbass2212,

Let me offer this method. It works all the time and is easy to understand.

I'll use your quadratic expression.

$4x^2-4x-15$

Multiply the coefficient of the quadratic term by the constant.

$4 \times -15 = -60$

You need to find 2 factors whose product is -60 and whose sum is -4 (the coefficient of the linear term).

It's not that hard. There is a small finite number of factors of -60.

$-2 \times 30$ Sum is not -4

$-3 \times 20$ Sum is not -4

$-4 \times 15$ Sum is not -4

$-5 \times 12$ Sum is not -4

$-6 \times 10$ Sum is not -4, but it is +4. So

$6 \times -10$ Sum is -4.

Now, it won't take you this many steps to come up with that combination. I just illustrated using the factors in sequential order. In reality, you should find them much quicker.

Now, what to do with these two factors. We will replace the linear coefficiet (-4) with these two factors. Watch:

$4x^2{\color{red}-10x+6x}-15$

Now use grouping to factor:

$(4x^2{\color{red}-10x)+(6x}-15)$

$2x(2x-5)+3(2x-5)$

$(2x+3)(2x-5)$

It seems like a long way, but it is very efficient.